Expressions with Absolute Value

Evaluate |a – 7| + 15 if a = 5.

|a – 7| + 15 = |5 – 7| + 15 Replace a with 5.

= |–2| + 15 5 – 7 = –2

= 2 + 15 |–2| = 2

= 17 Simplify.

Answer: 17

A. A

B. B

C. C

D. D A B C D

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A. 17

B. 24

C. 34

D. 46

Evaluate |17 – b| + 23 if b = 6.

Solve Absolute Value Equations

A. Solve |2x – 1| = 7. Then graph the solution set.

|2x – 1| = 7 Original equation

Case 1 Case 2

2x – 1 = 7 2x – 1 = –7

2x – 1 + 1 = 7 + 1 Add 1 to each side. 2x – 1 + 1 = –7 + 1

2x = 8 Simplify. 2x = –6

Divide each side by 2.

x = 4 Simplify. x = –3

Solve Absolute Value Equations

Answer: {–3, 4}

Solve Absolute Value Equations

B. Solve |p + 6| = –5. Then graph the solution set.

|p + 6| = –5 means the distance between p and 6 is –5. Since distance cannot be negative, the solution is the empty set Ø.

Answer: Ø

A. A

B. B

C. C

D. D A B C D

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A. Solve |2x + 3| = 5. Graph the solution set.

A. {1, –4}

B. {1, 4}

C. {–1, –4}

D. {–1, 4}

A. A

B. B

C. C

D. D A B C D

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B. Solve |x – 3| = –5.

A. {8, –2}

B. {–8, 2}

C. {8, 2}

D.

Solve an Absolute Value Equation

WEATHER The average January temperature in a northern Canadian city is 1°F. The actual January temperature for that city may be about 5°F warmer or colder. Write and solve an equation to find the maximum and minimum temperatures.

Method 1 Graphing

|t – 1| = 5 means that the distance between t and 1 is 5 units. To find t on the number line, start at 1 and move 5 units in either direction.

Solve an Absolute Value Equation

The solution set is {–4, 6}.

The distance from 1 to 6 is 5 units.

The distance from 1 to –4 is 5 units.

Method 2 Compound Sentence

Write |t –1| = 5 as t – 1 = 5 or t – 1 = –5.

Answer: The solution set is {–4, 6}. The range of temperatures is –4°F to 6°F.

Case 1 Case 2t – 1 = 5 t – 1 = –5

t – 1 + 1 = 5 + 1 Add 1 to each side.

t – 1 + 1 = –5 + 1

t = 6 Simplify. t = –4

Solve an Absolute Value Equation

A. A

B. B

C. C

D. D A B C D

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A. {–60, 60}

B. {0, 60}

C. {–45, 45}

D. {30, 60}

WEATHER The average temperature for Columbus on Tuesday was 45ºF. The actual temperature for anytime during the day may have actually varied from the average temperature by 15ºF. Solve to find the maximum and minimum temperatures.

Write an Absolute Value Equation

Write an equation involving absolute value for the graph.

Find the point that is the same distance from –4 as the distance from 6. The midpoint between –4 and 6 is 1.

Write an Absolute Value Equation

The distance from 1 to –4 is 5 units.

The distance from 1 to 6 is 5 units.

Answer: |y – 1| = 5

So, an equation is |y – 1| = 5.

A. A

B. B

C. C

D. D A B C D

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A. |x – 2| = 4

B. |x + 2| = 4

C. |x – 4| = 2

D. |x + 4| = 2

Write an equation involving the absolute value for the graph.